Next: About this document ...

Name e-mail
Rec.Teacher Day-hr

Exam 2 for 18.02, Spring 1999

For full credit try all problems. Write your name on each page and try if possible to do all your work on these pages. If it is necessary to add some more pages, write your name on each.

Problem 1 Consider the function near the point .

a)

Sketch the level curve of

through

$\begin{picture} (6,5)(-2,-2) \put(0,0){\vector(1,0){6}} \put(0,6.2){$y$} \put(-.... ...6,-.1){\line(0,1){.25}} \put(6.5,0){$x$} \put(1.5,3){\circle*{.3}} \end{picture}$

b)

Compute the gradient $\vec{\triangledown} w$ at

c)

Give an approximate formula for $\Delta w$ , the change in

, arising from small changes $\Delta x$ and $\Delta y$ in

and

starting from

d)

To changes in which of the variables

more sensitive at

e)

Find the directional derivative $\frac{d w}{ds} \vert _P$ in the direction of $\langle -1,1 \rangle$ .

Solution:

a): Accept any reasonable attempt here, it only needs to have the right derivative (or arguably so) at
b): $\vec{\triangledown} w=-3\frac{y^2}{x^4}{\bf i}+2\frac{y}{x^3}{\bf j}=-12{\bf i}+4{\bf j}$ at Accept vector in any identifiable form.
c): $\Delta w\simeq\vec{\triangledown}w\vert _P\cdot\Delta \vec{x}=-12\Delta x+4\Delta y.$
d): It is more sensitive to changes in Don't penalize earlier computational errors if the argument is correct.
e): $\frac{d w}{ds} \vert _{P,\hat e}=\vec{\triangledown}w\cdot\hat e=(12+4)/\sqrt 2=8\sqrt 2.$

Problem 2 Consider the surface defined by .

a)

Find the tangent plane at $\langle 1,1,2 \rangle$ --write it in the form

$\displaystyle ax+by+cz=d.$

b)

Find the point on the surface at which the normal has the same direction as ${\bf i} + 2 {\bf j} + \frac14{\bf k}$ .

Solution:

a): The gradient is $\vec{\triangledown}(xyz)=yz{\bf i}+xz{\bf j}+xy{\bf k}$ which is $2{\bf i}+2{\bf j}+{\bf k}$ at $\langle 1,1,2 \rangle.$ Thus the tangent plane is of the form that is
b): For the normal to have the same direction we must have $yz=\lambda,$ $xz=2\lambda,$ $xy=\frac14\lambda$ for some constant $\lambda.$ Since this implies $\frac12\lambda^3=4$ so $\lambda =2$ and $\langle x,y,z\rangle =\langle1,\frac12,4\rangle.$ This is the only point on the surface with the stated property.

Problem 3 Find all critical points of the function

$\displaystyle f(x,y)=4x^2+2y^2+3xy^2+1 \, .$

Work out the type (max/min/saddle/other) of each of them.

Solution: $\partial f/\partial x=8x+3y^2,$ $\partial f/\partial y=4y+6xy,$ $\partial^2 f/\partial x^2=8,$ $\partial^2f/\partial x\partial y=6y,$ $\partial^2f/\partial y^2=4+6x.$ Critical points therefore satisfy Thus either or The first gives The second gives so $y=\pm4/3.$ Thus there are three critical points, at and $(-2/3,\pm4/3).$ The discriminant, $\partial^2 f/\partial x^2\cdot\partial^2 f/\partial y^2-(\partial^2f/\partial x\partial y)^2$ at these points is respectively $8\cdot 4>0$ $8\cdot (4-4)-(\pm 4/3)^2<0$ so is a local minimum, and $(-2/3,\pm4/3)$ are both saddle points.

Problem 4 Suppose that and are (nice) functions of one variable. Show that the function of two variables

$\displaystyle z= f(x+y) +g(x-y)$

satisfies $\frac{\partial ^2 z}{\partial x^2} = \frac{\partial ^2z}{\partial y^2}$ (the wave equation).

Solution: By the chain rule, $\partial z/\partial x=f'(x+y)+g'(x-y),$ $\partial^2z/\partial x^2=f''(x+y)+g''(x-y)$ and similarly $\partial z/\partial y=f'(x+y)-g'(x-y)$ so $\partial^2z/\partial y^2=f''(x+y)+g''(x-y)=\partial^2z/\partial y^2$ as claimed.

Problem 5 Find the maximum and minimum values of the function on the ellipse .

Solution: Using Lagrange's method, find the critical points of $f(x,y)-\lambda g(x,y)$ with Thus $4+\lambda (4x)=0=2+\lambda (2y)$ so $x\lambda =-1=\lambda y.$ So $\lambda =0$ or The former case cannot occur, so hence $x=y=\pm1.$ At these two point and respectively, so the maximum is and the minimum is Maybe they can make substitution work, in which case it should be allowed.